Some experts have a hunch that for any nonnegative integer $r$ there are infinitely many elliptic curves over $\Bbb Q$ with Mordell-Weil rank at least $r$.

The best empirical evidence for this hunch can be found in Andrej Dujella's tables here and here, the strongest of this evidence being provided by Elkies using constructions involving K3 surfaces.

The only heuristic evidence I know of supporting this hunch is the results of Tate and Shafarevich (strengthened by Ulmer) that the Mordell-Weil ranks of elliptic curves over the a fixed function field $\Bbb F_q(t)$ can be arbitrarily large.

What other heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be unbounded?

Seeing as how some experts do no believe this conjecture, I'd also accept answer to the companion question:

What heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be uniformly bounded?

Cassels said it best (JLMS 1966, p.257):... it has been widely conjectured [on the basis of calculations] that there is an upper bound for the rank depending only on the groundfield. This seems to me implausible because the theory makes it clear that an abelian variety can only have high rank if it is defined by equations with very large coefficients. . (For there must be a lot of alternative factorizations to be possible in the arguments of §24.)
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JS MilneSep 11 '10 at 13:13

6 Answers
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Remke remarks: "I guess that the main evidence is the fact that every now and then someone comes up with a new record."

As a not-entirely-serious quantification of that remark, I graphed the following data: For each year starting in 1974 in which a new E(Q) rank record was found, graph the point (year,highest rank found that year). Here's the data: (1974,6),(1975,7),(1977,9),(1982,12),(1986,14),(1992,19), (1993,20),(1994,21),(1997,22),(1998,23),(2000,24),(2006,28),which I took from http://web.math.hr/~duje/tors/rankhist.html.
The data looks remarkably linear. The linear correlation coefficient is 0.998. The slope is 0.686. Some of the linearity probably comes from Moore's law (log-linear increase in computing power), but the table also represents major algorithmic advances of Mestre, Elkies, ..., so it's not clear (to me) why theoretical advances should also give linear growth!

I like this and find it a bit hilarious. This reminds me of a paper of Arjen Lenstra's about selecting cryptographic key sizes in which he postulates a kind of Moore's law for mathematical advances.
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Victor MillerJan 1 '11 at 23:20

There are dozens of articles in the mathematical literature giving a rather strong correlation between the ranks of certain elliptic curves (for example $y^2 = k^3 + k$ for certain values of $k$) and the $3$-rank of the class groups of quadratic number fields (Quer, Top, DeLong), or the $2$-rank of the class groups of pure cubic fields (Frey et al, Schneiders). By Lenstra-Cohen, we certainly expect the former to be unbounded; I don't know whether Cohen-Martinet predicts that the second set is not bounded since $2$ is some kind of bad prime in this connection. I do wonder, however, whether there really are (not were!) experts who suggest that these ranks are bounded.

I have also seen attempts at transferring the Cohen-Lenstra heuristics to the Tate-Shafarevich group of elliptic curves (C. Delaunay); I do not know whether these heuristics are strong enough to predict that sha is small often enough to give a large rank of the Mordell-Weil group.

I assume you mean $y^2 = x^3 + k$ in your parenthetical remark. As far as whether there are (not were!) experts who believe that ranks might be bounded. I realize that I can't date the thirdhand story I heard that made me change the original phrasing of my question.
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Jamie WeigandtSep 14 '10 at 3:23

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Delaunay's heuristics describe the order of Sha as the curves vary in a family of elliptic curve of fixed rank. In fact, the probability that Sha has a certain order depends on the rank in this model. So these heuristics won't say anything about the rank.
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Chris WuthrichSep 14 '10 at 13:05

I don't think I believe this anymore, Class groups are related to Selmer groups and Sha, which we know can be large, but they rank wouldn't have to be large. There definitely are experts who believe ranks could be bounded. I should describe some of those heuristics (Granville-Watkins) and make is community wiki when I get a chance.
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Jamie WeigandtApr 11 '14 at 18:11

I am not aware of much evidence for arbitrary high rank elliptic curves. Silverman in his book (Arithmetic of elliptic curves) gives as evidence the lack of evidence for the opposite statement. I guess that the main evidence is the fact that every now and then someone comes up with a new record.

The function field case is some sort of evidence, however, most constructions rely on supersingular elliptic surfaces or supersingular Fermat surfaces. These objects are quite far away from characteristic zero objects. (One exception to this is the construction by Bouw-Diem-Scholten.)
Shioda tried to use Fermat surfaces in order to produce high rank elliptic curves over $K(t)$, with $char(K)=0$, but the best you can get from that is rank 68 if $K$ is algebraically closed, for $K=\mathbb{Q}$ this maximum is much lower.

A second piece of evidence is the fact that we can construct big Selmer groups. We can find arbitrarily many elements of order m in the m-Selmer group, for $m=2,\dots,10,12,13,16,25$. However, every construction of large Selmer groups seems to allow a refinement that makes the Tate-Shafarevich group large. (For some reason nobody wrote this up for the composite $m$ mentioned above, but I am quite sure that this has been done.)
Matsuno gave a construction s.t. for each prime number $p$ and each field $K$ containing a degree $p$ Galois extension of $\mathbb{Q}$ one can produce a series of elliptic curves $E/\mathbb{Q}$ the group $S^p(E/K)$ can be arbitrarily large, but also this construction allows a refinement to get big Tate-Shafarevich groups.

Dear Felipe,
as long as we restrict ourselves to elliptic curves over $\mathbb{Q}$, the only results known in this direction are that the 2, 3, 5, 7 and 13 primary torsion of Tate-Shafarevich groups can be arbitrarily large. These are due to Kramer, Cassels, Fisher, Matsuno and Matsuno, respectively. It is still unknown whether it is true that for any $p$, the $p$ primary torsion of sha can be arbitrarily large. It is not even known whether the $p$-torsion of sha can be non-trivial for arbitrary $p$.

As far as I know, almost everybody nowadays believes that ranks of elliptic curves over $\mathbb{Q}$ can be arbitrarily large, but when pressed for evidence, most people point to the function field case. In fact, it is not even known, whether the rank can get arbitrarily large over number fields of uniformly bounded degree.

I was thinking of the easier fact that you can make the Selmer group for the p-descent big by making the curve have many places of multiplicative reduction, so either the rank is large or the p-torsion in Sha is large. This should work for all p. Disentangling whether the contribution comes from rank or Sha is another matter.
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Felipe VolochSep 13 '10 at 20:04

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I am pretty sure that this won't be quite so easy for arbitrary $p$. As Remke mentions in his answer, arbitrarily large $p$-Selmer groups over $\mathbb{Q}$ are only known to exist for a few specific values of $p$ and more or less each one of them has been worth a publication.
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Alex B.Sep 14 '10 at 0:27

It is generally believed that a "random" elliptic curve over $\mathbb Q$ will have rank 0 or 1, just depending on the sign of the functional equation, and hence that the average rank is 1/2. But everyone I know who believes this also believes in the unboundedness of the rank; it is just that high rank elliptic curves will be very rare.
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EmertonSep 11 '10 at 0:03

I thought about this. There's a conjecture of Goldfeld and Szpiro, which is (assuming BSD) equivalent to the ABC conjecture which bounds the order of sha in terms of the conductor. These elliptic curves with large Selmer groups are likely to have very large conductors so this wouldn't force large rank.
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Jamie WeigandtSep 12 '10 at 4:44